randRange(0,2)
randFromArray([
["O", "A", "B", "C"],
["B", "A", "C", "D"],
["O", "L", "M", "N"],
["P", "Q", "R", "S"]
])
"\\angle " + A + O + B
"\\angle " + B + O + C
"\\angle " + A + O + C
shuffle([ ANGLE_BOT, ANGLE_TOP ])
(function() {
if ( PROBLEM === 0 ) {
return randRange(30, 170)
}
else if ( PROBLEM === 1) {
return 180
}
else {
return 90;
}
})()
randRange(15, largeAngle-15)
largeAngle - subAngle1
randRange( 1, 20 )
randRange( 2, 9 )
subAngle1 - COEF_1*X
randRange( 2, 9 )
subAngle2 - COEF_2*X
randFromArray([ [ANGLE_ONE, COEF_1, CONST_1], [ANGLE_TWO, COEF_2, CONST_2] ])
Given
\qquad \overrightarrow{OA}\perp\overrightarrow{OC},
\qquad m ANGLE_ONE = COEF_1x + CONST_1, and
\qquad m ANGLE_TWO = COEF_2x + CONST_2,
, and
\qquad m LARGE_ANGLE = largeAngle,
find mANSWER[0].
init({
range: [ [(Math.min( polar(5.5, largeAngle)[0], -.5)-.5), 5.5], [-1, (Math.min( Math.max( polar(5.5,largeAngle)[1], polar(5.5, subAngle2)[1])+.5, 5.5))] ],
scale: 40
});
dotAttr={ r: 0.1, fill: "black", stroke: "none" };
line( [0, 0], [5, 0], { arrows: "->" } );
line( [0, 0], polar(5, largeAngle), {arrows: "->"} );
if (ANGLE_ONE === ANGLE_TOP){
line( [0, 0], polar(5, subAngle2), {arrows: "->"} );
} else {
line( [0, 0], polar(5, subAngle1), {arrows: "->"} );
}
circle( [0, 0] , dotAttr );
label( [0, 0], O, "below left" );
circle( polar(4.5,largeAngle), dotAttr );
label( polar(4.5,largeAngle), A, (largeAngle > 120) ? "below left" : "left" );
circle( [4.5, 0], dotAttr );
label( [4.5, 0], C, "below" );
if (ANGLE_ONE === ANGLE_TOP){
circle( polar(4.5, subAngle2), dotAttr );
label( polar(4.5, subAngle2), B, (subAngle2 > 45) ? "right" : "above" );
} else {
circle( polar(4.5, subAngle1), dotAttr );
label( polar(4.5, subAngle1), B, (subAngle1 > 45) ? "right" : "above" );
}
ANSWER[1] * X + ANSWER[2]
{}^{\circ}
From the diagram, we see that together \blue{ANGLE_TOP} and \green{ANGLE_BOT} form \purple{LARGE_ANGLE}, so
\qquad \blue{mANGLE_TOP} + \green{mANGLE_BOT} = \purple{mLARGE_ANGLE}.
style({ stroke: GREEN, strokeWidth: 3 });
arc([0, 0], 1, 0, ANGLE_ONE === ANGLE_TOP ? subAngle2 : subAngle1).toBack();
style({ stroke: BLUE, strokeWidth: 3 });
arc([0, 0], 1, ANGLE_ONE === ANGLE_TOP ? subAngle2 : subAngle1, largeAngle).toBack();
if (PROBLEM === 0){
style({ stroke: "purple", strokeWidth: 3 });
arc( [ 0, 0 ], 1.2, 0, largeAngle );
jQuery( "#givenLarge" ).css({ "color": "purple"} );
}
if (ANGLE_ONE === ANGLE_TOP){
jQuery( "#given0" ).css({ "color": BLUE });
jQuery( "#given1" ).css({ "color": GREEN });
} else {
jQuery( "#given0" ).css({ "color": GREEN });
jQuery( "#given1" ).css({ "color": BLUE });
};
Since LARGE_ANGLE is a straight angle, we know \purple{mLARGE_ANGLE = 180}.
style({ stroke: "purple", strokeWidth: 3 });
arc( [ 0, 0 ], 1.2, 0, 180 );
Since we are given that \overrightarrow{OA}\perp\overrightarrow{OC}, we know \purple{mLARGE_ANGLE = 90}.
style({ stroke: "purple", strokeWidth: 3 });
path( [[0, .5], [.5, .5], [.5, 0]] );
Substitute in the expressions that were given for each measure:
\qquad \blue{COEF_1x + CONST_1} + \green{COEF_2x + CONST_2} = \purple{largeAngle}.
\qquad \blue{COEF_2x + CONST_2} + \green{COEF_1x + CONST_1} = \purple{largeAngle}.
Combine like terms:
\qquadCOEF_1 + COEF_2x + CONST_1 + CONST_2 = largeAngle.
Subtract CONST_1 + CONST_2 from both sides:
Add -(CONST_1 + CONST_2) to both sides:
\qquadCOEF_1 + COEF_2x = largeAngle - CONST_1 - CONST_2.
Divide both sides by COEF_1 + COEF_2 to find x:
\qquad x = X.
Substitute X for x in the expression that was given for mANSWER[0]:
\qquad mANSWER[0] = ANSWER[1](\pink{X}) + ANSWER[2].
Substitute X for x in the expression that was given for mANSWER[0]:
\qquad mANSWER[0] = ANSWER[1](\pink{X}) + ANSWER[2].
Simplify:
\qquad \green{mANSWER[0] = ANSWER[1]*X + ANSWER[2]}.
Simplify:
\qquad \blue{mANSWER[0] = ANSWER[1]*X + ANSWER[2]}.
So \green{mANSWER[0] = ANSWER[1] * X + ANSWER[2]}.
So \blue{mANSWER[0] = ANSWER[1] * X + ANSWER[2]}.